2 Echelon FormsThis matrix which have following properties is in reduced row-echelon form (Example 1, 2).1. If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1. We call this a leader 1.2. If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix.3. In any two successive rows that do not consist entirely of zeros, the leader 1 in the lower row occurs farther to the right than the leader 1 in the higher row.4. Each column that contains a leader 1 has zeros everywhere else.A matrix that has the first three properties is said to be in row-echelon form (Example 1, 2).A matrix in reduced row-echelon form is of necessity in row-echelon form, but not conversely.
3 Example 1 Row-Echelon & Reduced Row-Echelon form
4 Example 2 More on Row-Echelon and Reduced Row-Echelon form All matrices of the following types are in row-echelon form ( any real numbers substituted for the *’s. ) :All matrices of the following types are in reduced row-echelon form ( any real numbers substituted for the *’s. ) :
5 Example 3 Solutions of Four Linear Systems (a) Suppose that the augmented matrix for a system of linear equations have been reduced by row operations to the given reduced row-echelon form. Solve the system.Solution (a)the corresponding system of equations is :
6 Example 3 Solutions of Four Linear Systems (b1) Solution (b)1. The corresponding system of equations is :free variablesleading variables
7 Example 3 Solutions of Four Linear Systems (b2) 2. We see that the free variable can be assigned an arbitrary value, say t, which then determines values of the leading variables.3. There are infinitely many solutions, and the general solution is given by the formulas
8 Example 3 Solutions of Four Linear Systems (c1) Solution (c)The 4th row of zeros leads to the equation places no restrictions on the solutions (why?). Thus, we can omit this equation.
9 Example 3 Solutions of Four Linear Systems (c2) Solution (c)Solving for the leading variables in terms of the free variables:The free variable can be assigned an arbitrary value,there are infinitely many solutions, and the general solution is given by the formulas.
10 Example 3 Solutions of Four Linear Systems (d) Solution (d):the last equation in the corresponding system of equation isSince this equation cannot be satisfied, there is no solution to the system.
11 Elimination Methods (1/7) We shall give a step-by-step elimination procedure that can be used to reduce any matrix to reduced row-echelon form.
12 Elimination Methods (2/7) Step1. Locate the leftmost column that does not consist entirely of zeros.Step2. Interchange the top row with another row, to bring a nonzero entry to top of the column found in Step1.Leftmost nonzero columnThe 1th and 2th rows in the preceding matrix were interchanged.
13 Elimination Methods (3/7) Step3. If the entry that is now at the top of the column found in Step1 is a, multiply the first row by 1/a in order to introduce a leading 1.Step4. Add suitable multiples of the top row to the rows below so that all entires below the leading 1 become zeros.The 1st row of the preceding matrix was multiplied by 1/2.-2 times the 1st row of the preceding matrix was added to the 3rd row.
14 Elimination Methods (4/7) Step5. Now cover the top row in the matrix and begin again with Step1 applied to the submatrix that remains. Continue in this way until the entire matrix is in row-echelon form.Leftmost nonzero column in the submatrixThe 1st row in the submatrix was multiplied by -1/2 to introduce a leading 1.
15 Elimination Methods (5/7) Step5 (cont.)-5 times the 1st row of the submatrix was added to the 2nd row of the submatrix to introduce a zero below the leading 1.The top row in the submatrix was covered, and we returned again Step1.Leftmost nonzero column in the new submatrixThe first (and only) row in the new submetrix was multiplied by 2 to introduce a leading 1.The entire matrix is now in row-echelon form.
16 Elimination Methods (6/7) Step6. Beginning with las nonzero row and working upward, add suitable multiples of each row to the rows above to introduce zeros above the leading 1’s.7/2 times the 3rd row of the preceding matrix was added to the 2nd row.-6 times the 3rd row was added to the 1st row.5 times the 2nd row was added to the 1st row.The last matrix is in reduced row-echelon form.
17 Elimination Methods (7/7) Step1~Step5: the above procedure produces a row-echelon form and is called Gaussian elimination.Step1~Step6: the above procedure produces a reduced row-echelon form and is called Gaussian-Jordan elimination.Every matrix has a unique reduced row-echelon form but a row-echelon form of a given matrix is not unique.
18 Example 4 Gauss-Jordan Elimination(1/4) Solve by Gauss-Jordan EliminationSolution:The augmented matrix for the system is
19 Example 4 Gauss-Jordan Elimination(2/4) Adding -2 times the 1st row to the 2nd and 4th rows givesMultiplying the 2nd row by -1 and then adding -5 times the new 2nd row to the 3rd row and -4 times the new 2nd row to the 4th row gives
20 Example 4 Gauss-Jordan Elimination(3/4) Interchanging the 3rd and 4th rows and then multiplying the 3rd row of the resulting matrix by 1/6 gives the row-echelon form.Adding -3 times the 3rd row to the 2nd row and then adding 2 times the 2nd row of the resulting matrix to the 1st row yields the reduced row-echelon form.
21 Example 4 Gauss-Jordan Elimination(4/4) The corresponding system of equations isSolutionThe augmented matrix for the system isWe assign the free variables, and the general solution is given by the formulas:
22 Back-SubstitutionIt is sometimes preferable to solve a system of linear equations by using Gaussian elimination to bring the augmented matrix into row-echelon form without continuing all the way to the reduced row-echelon form.When this is done, the corresponding system of equations can be solved by solved by a technique called back-substitution.Example 5
23 Example 5 ex4 solved by Back-substitution(1/2) From the computations in Example 4, a row-echelon form from the augmented matrix isTo solve the corresponding system of equationsStep1. Solve the equations for the leading variables.
24 Example5 ex4 solved by Back-substitution(2/2) Step2. Beginning with the bottom equation and working upward, successively substitute each equation into all the equations above it.Substituting x6=1/3 into the 2nd equationSubstituting x3=-2 x4 into the 1st equationStep3. Assign free variables, the general solution is given by the formulas.
25 Example 6 Gaussian elimination(1/2) Solve by Gaussian elimination andback-substitution. (ex3 of Section1.1)SolutionWe convert the augmented matrixto the ow-echelon formThe system corresponding to this matrix is
26 Example 6 Gaussian elimination(2/2) SolutionSolving for the leading variablesSubstituting the bottom equation into those aboveSubstituting the 2nd equation into the top
27 Homogeneous Linear Systems(1/2) A system of linear equations is saidto be homogeneous if the constantterms are all zero; that is , thesystem has the form :Every homogeneous system of linear equation is consistent, since all such system haveas a solution. This solution is called the trivial solution; if there are another solutions, they are called nontrivial solutions.There are only two possibilities for its solutions:The system has only the trivial solution.The system has infinitely many solutions in addition to the trivial solution.
28 Homogeneous Linear Systems(2/2) In a special case of a homogeneous linear system of two linear equations in two unknowns: (fig1.2.1)
29 Example 7 Gauss-Jordan Elimination(1/3) Solve the following homogeneous system of linear equations by using Gauss-Jordan elimination.SolutionThe augmented matrixReducing this matrix to reduced row-echelon form
30 Example 7 Gauss-Jordan Elimination(2/3) Solution (cont)The corresponding system of equationSolving for the leading variables isThus the general solution isNote: the trivial solution is obtained when s=t=0.
31 Example7 Gauss-Jordan Elimination(3/3) Two important points:Non of the three row operations alters the final column of zeros, so the system of equations corresponding to the reduced row-echelon form of the augmented matrix must also be a homogeneous system.If the given homogeneous system has m equations in n unknowns with m<n, and there are r nonzero rows in reduced row-echelon form of the augmented matrix, we will have r<n. It will have the form:
32 Theorem 1.2.1A homogeneous system of linear equations with more unknowns than equations has infinitely many solutions.Note: Theorem applies only to homogeneous systemExample 7 (3/3)
33 Computer Solution of Linear System Most computer algorithms for solving large linear systems are based on Gaussian elimination or Gauss-Jordan elimination.IssuesReducing roundoff errorsMinimizing the use of computer memory spaceSolving the system with maximum speed
34 Exercise Set 1.2 Question 11Solve the following system by Gaussian elimination